Question

In the figure shown below, force in member AB will be _________ kN.

• A. 198

Answer 1

The correct option is A 198
By method of sections,

$\Sigma M_G=0;\Rightarrow F_{AB}\cos 45{}^{\circ }\times 2+F_{AB}\sin 45{}^{\circ }\times 2+40\times 2+80\times 6=0$

$\Rightarrow \sqrt{2}{F}_{AB}+F_{AB}\sqrt{2}+80+480=0$

$\Rightarrow F_{AB}=-197.989kN\simeq 197.99kN\simeq 198kN\left(compressive\right)$